CHAPTER 9 Vector Calculus

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CHAPTER 9Vector Calculus
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Contents

• 9.5 Directional Derivatives
• 9.6 Tangent Planes and Normal Lines
• 9.7 Divergence and Curl
• 9.8 Lines Integrals
• 9.9 Independence of Path

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9.5 Directional Derivative

• IntroductionSee Fig 9.26.

4
The Gradient of a Function

• Define the vector differential operator
  asthen (1) (2)are the
  gradients of the functions.

5
Example 1

• Compute
• Solution

6
Example 2

• If F(x, y, z) xy2 3x2 z3, find the gradient
  at (2, 1, 4).
• Solution

7
DEFINITION 9.5
Directional Derivatives

• The directional derivative of z f(x, y) in the
  direction
• of a unit vector u cos ?i sin ?j
  is (4)provided the limit exists.

8
Fig 9.27
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THEOREM 9.6
If z f(x, y) is a differentiable function of
x and y, and u cos ?i sin ?j,
then (5)
Computing a Directional Derivative

• ProofLet x, y and ? be fixed, then g(t) f(x
  t cos ?, y t sin ?) is a function of one
  variable.

10

• First
• Second by chain rule

11

• Here the subscripts 1 and 2 refer to partial
  derivatives of f(x t cos ?, y t sin ?)
  w.s.t. (x t cos ?) and (y t sin ?). When t
  0, x t cos ? and y t sin ? are simply x and
  y, then (7) becomes (8)
• Comparing (4), (6), (8), we have

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Example 3

• Find the directional derivative of f(x, y)
  2x2y3 6xy at (1, 1) in the direction of a unit
  vector whose angle with the positive x-axis is
  ?/6.
• Solution

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Example 3 (2)

• Now, ? ?/6, u cos ?i sin ?j becomesThen

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Example 4

• Consider the plane perpendicular to xy-plane and
  passes through P(2, 1), Q(3, 2). What is the
  slope of the tangent line to the curve of
  intersection of this plane and f(x, y) 4x2 y2
  at (2, 1, 17) in the direction of Q.
• SolutionWe want Duf(2, 1) in the direction
  given by , and form a unit
  vector

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Example 4 (2)

• Nowthen the requested slope is

16
Functions of Three Variables

• where ?, ?, ? are the direction angles of
  the vector u measured relative to the positive x,
  y, z axis. But as before, we can show
  that (9)

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• Since u is a unit vector, from (10) in Sec 7.3
  thatIn addition, (9) shows

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Example 5

• Find the directional derivative of F(x, y, z)
  xy2 4x2y z2 at (1, 1, 2) in the direction 6i
  2j 3k.
• SolutionSincewe have

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Example 5 (2)

• Since ?6i 2j 3k? 7, then u (6/7)i (2/7)j
  (3/7)k is a unit vector. It follows from (9)
  that

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Maximum Value of the Direction Derivative

• From the fact thatwhere ? is the angle between
  and u. Becausethen

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• In other words, The maximum value of the
  direction derivative is and it occurs
  when u has the same direction as (when cos ?
  1) , (10)and The minimum value of the
  direction derivative is and it occurs
  when u has opposite direction as (when cos ?
  -1) (11)

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Example 6

• In Example 5, the maximum value of the
  directional derivative at (1, -1, 2) is and
  the minimum value is .

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Gradient points in Direction of Most Rapid
Increase of f

• Put another way, (10) and (11) stateThe gradient
  vector points in the direction in which f
  increase most rapidly, whereas points in
  the direction of the most rapid decrease of f.

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Example 7

• Each year in L.A. there is a bicycle race up to
  the top of a hill by a road known to be the
  steepest in the city. To understand why a
  bicyclist with a modicum of sanity will zigzag up
  the road, let us suppose the graph ofshown in
  Fig 9.28(a) is a mathematical model of the hill.
  The gradient of f is

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Example 7 (2)

• where r xi yj is a vector pointing to the
  center of circular base. Thus the steepest
  ascent up the hill is a straight road whose
  projection in the xy-plane is a radius of the
  circular base. Since a bicyclist will zigzag a
  direction u other than to reduce this
  component. See Fig 9.28.

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Fig 9.28
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Example 8

• The temperature in a rectangular box is
  approximated by If a mosquito is located at (
  ½, 1, 1), in which the direction should it fly up
  to cool off as rapidly as possible?

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Example 8 (2)

• SolutionThe gradient of T is
• Therefore, To cool off most rapidly, it should
  fly in the direction -¼k, that is, it should dive
  for the floor of the box, where the temperature
  is T(x, y, 0) 0

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9.6 Tangent Plane and Normal Lines

• Geometric Interpretation of the Gradient
  Functions of Two VariablesSuppose f(x, y) c is
  the level curve of z f(x, y) passes through
  P(x0, y0), that is, f(x0, y0) c.If x g(t), y
  h(t) such that x0 g(t0), y0 h(t0), then the
  derivative of f w.s.t. t is (1)When
  we introduce

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• then (1) becomes When at t t0, we
  have (2)Thus, if ,
  is orthogonal toat P(x0, y0). See
  Fig 9.30.

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Fig 9.30
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Example 1

• Find the level curves of f(x, y) -x2 y2
  passing through (2, 3). Graph the gradient at the
  point.
• Solution Since f(2, 3) 5, we have -x2 y2
  5.NowSee Fig 9.31.

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Fig 9.31
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Geometric Interpretation of the Gradient
Functions of Three Variables

• Similar concepts to two variables, the derivative
  of F(f(t), g(t), h(t)) c implies
  (3)In particular, at t t0, (3) is
  (4)See Fig 9.32.

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Fig 9.32
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Example 2

• Find the level surfaces of F(x, y, z) x2 y2
  z2 passing through (1, 1, 1). Graph the gradient
  at the point.
• Solution Since F(1, 1, 1) 3,then x2 y2 z2
  3See Fig 9.33.

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Fig 9.33
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DEFINITION 9.6
Tangent Plane

• Let P(x0, y0, z0) is a point on the graph of F(x,
  y, z) c,
• where ?F is not 0. The tangent plane at P is a
  plane
• through P and is perpendicular to ?F evaluated at
  P.

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THEOREM 2.1
Let P(x0, y0, z0) is a point on the graph of
F(x, y, z) c, where ?F is not 0. Then an
equation of this tangent plane at P is Fx(x0,
y0, z0)(x x0) Fy(x0, y0, z0)(y y0)
Fz(x0, y0, z0)(z z0) 0 (5)
Criterion for an Extra Differential

• That is, . See Fig 9.34.

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Fig 9.34
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Example 3

• Find the equation of the tangent plane to x2
  4y2 z2 16 at (2, 1, 4).
• SolutionF(2, 1, 4) 16, the did graph passes
  (2, 1, 4). Now Fx(x, y, z) 2x, Fy(x, y, z)
  8y, Fz(x, y, z) 2z, thenFrom (5) we have the
  equation 4(x 2) 8(y 1) 8(z 4) 0 or
  x 2y 2z 8.

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Surfaces Given by z f(x, y)

• When the equation is given by z f(x, y), then
  we can set F z f(x, y) or F f(x, y) z.

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Example 4

• Find the equation of the tangent plane to z ½x2
  ½ y2 4 at (1, 1, 5).
• SolutionLet F(x, y, z) ½x2 ½ y2 z 4.
  This graph did pass (1, 1, 5), since F(1, 1, 5)
  0. Now Fx x, Fy y, Fz 1, then
• From (5), the desired equation is (x 1)
  (y 1) (z 5) 0 or x y z 7

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Fig 9.35
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Normal Line

• Let P(x0, y0, z0) is on the graph of F(x, y, z)
  c, where ?F ? 0. The line containing P that is
  parallel to ?F(x0, y0, z0) is called the normal
  line to the surface at P.

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Example 5

• Find parametric equations for the normal line to
  the surface in Example 4 at (1, 1, 5).
• SolutionA direction vector for the normal line
  at (1, 1, 5) is ?F(1, 1, 5) i j kthen
  the desired equations are x 1 t, y 1
  t, z 5 t

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9.7 Divergence and Curl

• Vector Functions F(x, y) P(x, y)i Q(x,
  y)j F(x, y, z) P(x, y, z)i Q(x, y, z)j
  R(x, y, z)k

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Fig 9.37 (a) (b)
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Fig 9.37 (c) (d)
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Example 1

• Graph F(x, y) yi xj
• SolutionSinceletFor and
  k 2, we have(i) x2 y2 1at (1, 0), (0,
  1), (1, 0), (0, 1), the corresponding vectors
  j, i, j , i have the same length 1.

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Example 1 (2)

• (ii) x2 y2 2at (1, 1), (1, 1), (1, 1),
  (1, 1), the corresponding vectors i j, i
  j, i j, i j have the same length .
• (iii) x2 y2 4at (2, 0), (0, 2), (2, 0), (0,
  2), the corresponding vectors 2j, 2i, 2j, 2i
  have the same length 2. See Fig 9.38.

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Example 1 (3)
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DEFINITION 9.7
Curl
The curl of a vector field F Pi Qj Rk is
the vector field

• In practice, we usually use this
  form (1)

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DEFINITION 9.8
Divergence
The divergence of a vector field F Pi Qj
Rk is the scalar function

• Observe that we can also use this
  form (4)

55
Example 2

• If F (x2y3 z4)i 4x5y2zj y4z6 k, find curl
  F and div F?
• Solution

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Example 2 (2)
57
Please prove

• If f is a scalar function with continuous second
  partial derivatives, then (5)If F is
  a vector field with continuous second partial
  derivatives, then (6)

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Physical Interpretations

• See Fig 9.41. The curl F is a measure of tendency
  of the fluid to turn the device about its
  vertical axis w. If curl F 0, then the flow of
  the fluid is said to be irrotational. Also see
  Fig 9.42.

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Fig 9.41
60
Fig 9.42
61

• From definition 9.8 we saw that div F near a
  point is the flux per unit volume. (i) If div
  F(P) gt 0 source for F.(ii) If div F(P) lt 0
  sink for F.(iii) If div F(P) 0 no sources or
  sinks near P.Besides, If ? ? F 0
  incompressible or solenoidal.See Fig 9.43.

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Fig 9.43

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9.8 Line Integrals

• TerminologyIn Fig 9.46, we show four new
  terminologies.
• Fig 9.46.

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Line Integral in the Plane

• If C is a smooth curve defined by x f(t), y
  g(t), a ? t ? b. Since dx f(t) dt, dy g(t)
  dt, and (which is called
  the differential of arc length), then we have
  (1) (2) (3)

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Example 1

• Evaluate (a) (b)
  (c)on the ¼ circle C defined by Fig 9.47

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Example 1 (2)

• Solution(a)

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Example 1 (3)

• (b)

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Example 1 (4)

• (c)

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Method of Evaluation

• If the curve C is defined by y f(x), a ? x ? b,
  then we have dy f (x) dx andThus
  (4) (5) (6)
• Note If C is composed of two smooth curves C1
  and C2, then

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• Notation In many applications, we havewe
  usually write as or (7)A line
  integral along a close curve

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Example 2

• Evaluate where C
• Solution See Fig 9.48. Using dy 3x2 dx,

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Fig 9.48
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Example 3

• Evaluate , where C
• Solution

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Example 4

• Evaluate , where C is
  shown in Fig 9.49(a).
• Solution Since C is piecewise smooth, we express
  the integral as See Fig 9.49(b).

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Fig 9.49
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Example 4 (2)

• (i) On C1, we use x as a parameter. Since y 0,
  dy 0,
• (ii) On C2, we use y as a parameter. Since x 2,
  dx 0,

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Example 4 (3)

• (iii) On C3, we use x as a parameter. Since y
  x2, dy 2x dx, Hence,

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• Note See Fig 9.50, where -C denote the curve
  having opposite orientation, then
  Equivalently, (8)For example, in
  (a) of Example 1,

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Fig 9.50
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Lines Integrals in Space
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Method to Evaluate Line Integral in Space

• If C is defined by then we haveSimilar
  method can be used for

82

• andWe usually use the following form

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Example 5

• Evaluate , where C is
• Solution Since we havewe get

84
Another Notation

• Let r(t) f(t)i g(t)j, then dr(t)/dt
  f(t)i g(t)j (dx/dt)i (dy/dt)jNow if
  F(x, y) P(x, y)i Q(x, y)jthus (10)Wh
  en on a space (11)where F(x, y, z)
  Pi Qj Rk dr dx i dy j dz k,

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Work

• If A and B are the points (f(a), g(a)) and (f(b),
  g(b)). Suppose C is divided into n subarcs of
  lengths ?sk. On each subarc F(xk, yk) is a
  constant force. See Fig 5.91(a).
• If, as shown in Figure 9.51(b), the length of the
  vector is an approximation to the length of the
  kth subarc, then the approximate work done by F
  over the subarc is

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• The work done by F along C is as the line
  integral or (12)Sincewe let dr Tds,
  where T dr / ds is a unit tangent to
  C. (13)The work done by a force F along
  a curve C is due entirely to the tangential
  component of F.

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Fig 9.51
88
Example 6

• Find the work done by (a) F(x, y) xi yj(b) F
  (¾ i ½ j) along the curve C traced by r(t)
  cos ti sin tj, from t 0 to t ?.
• Solution (a) dr(t) (- sin ti cos tj) dt,
  then

89
Fig 9.52
90
Example 6 (2)

• (b) See Fig 9.53.

91
Fig 9.53

92
Circulation

• Note Let dr T ds, where T dr / ds, then
• Circulation of F around C

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Fig 9.54
94
9.9 Independence of Path

• Differential For two variablesFor three
  variables

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Path Independence

• A line integral whose value is the same for every
  curve or path connecting A and B.

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Example 1

• has the same value on
  each path between (0, 0) and (1, 1) shown in Fig
  9.65. Recall that

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Fig 9.65
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THEOREM 9.8
Suppose there exists a ?(x, y) such that d?
Pdx Qdy, that is, Pdx Qdy is an exact
differential. Then
depends on only the endpoints A and B, and
Fundamental Theorem for Line Integrals
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THEOREM 9.8 Proof

• Let C be a smooth curve The endpoints are
  (f(a), g(a)) and (f(b), g(b)), then

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Two facts

• (i) This is also valid for piecewise smooth
  curves.(ii) The converse of this theorem is also
  true. is independent of path
  iff P dx Q dy is an exact differential. (1)
• Notation for a line integral independent of path
  

101
Example 2

• Since d(xy) y dx x dy, y dx x dy is an
  exact differential. Hence
  is independent of path. Especially, if the
  endpoints are (0, 0) and (1, 1), we have

102
Simply Connected Region in the Plane

• Refer to Fig 9.66. Besides, a simply connected
  region is open if it contains no boundary points.

103
Fig 9.66
104
THEOREM 9.9
Let P and Q have continuous first partial
derivatives in an open simply connected region.
Then is independent of the path C if and only
iffor all (x, y) in the region.
Test for Path Independence
105
Example 3

• Show that is not independent of path C.
• SolutionWe have P x2 2y3 and Q x 5y,
  then andSince , we
  complete the proof.

106
Example 4

• Show that is independent of any path between
  (-1, 0) and (3, 4). Evaluate.
• SolutionWe have P y2 6xy 6 and Q 2xy
  3x2, then and This is an exact
  differential.

107
Example 4 (2)

• Suppose there exists a ? such that Integrating
  the first, we have then
  we have , g(y) C.

108
Example 4 (3)

• Since d(y2x 3x2y 6x C) d(y2x 3x2y
  6x), we simply usethen

109
Another Approach

• We know y x 1 is one of the paths connecting
  (-1, 0) and (3, 4). Then

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Conservative Vector Fields

• If a vector field is independent of path, we
  havewhere F Pi Qj is a vector field and
• In other words, F is the gradient of ?. Since F
  ??, F is said to be a gradient field and the ? is
  called the potential function of F. Besides, we
  all call this kind of vector field to be
  conservative.

111
Example 5

• Show that F (y2 5)i (2xy 8)j is a
  gradient field. Find a potential function for F.
• SolutionSince thenwe have

112
THEOREM 9.10
Let P, Q, and R have continuous first partial
derivatives in an open simply connected region of
space. Then is
independent of the path C if and only if
Test for Path Independence
113
Example 6

• Show that is independent of path between (1, 1,
  1) and (2, 1, 4).
• SolutionSinceit is independent of path.

114
Example 6 (2)

• Suppose there exists a function ?, such
  thatIntegrating the first w.s.t. x, then It
  is the factthus

115
Example 6 (3)

• Nowand we have and
  h(z) z C. Disregarding C, we
  get (2)

116
Example 6 (4)

• Finally,

117

• From example 6, we know that F is a conservative
  vector field, and can be written as F ??.
  Remember in Sec 9.7, we have ???? 0, thus
  F is a conservative vector field iff ?
  ? F 0

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Thank You for Your Attention.